What is the extraneous solution to these equations? $\dfrac{x^2 + 2}{x - 2} = \dfrac{6x + 18}{x - 2}$
Multiply both sides by $x - 2$ $ \dfrac{x^2 + 2}{x - 2} (x - 2) = \dfrac{6x + 18}{x - 2} (x - 2)$ $ x^2 + 2 = 6x + 18$ Subtract $6x + 18$ from both sides: $ x^2 + 2 - (6x + 18) = 6x + 18 - (6x + 18)$ $ x^2 + 2 - 6x - 18 = 0$ $ x^2 - 16 - 6x = 0$ Factor the expression: $ (x - 8)(x + 2) = 0$ Therefore $x = 8$ or $x = -2$ The original expression is defined at $x = 8$ and $x = -2$, so there are no extraneous solutions.